hdu 2586 How far away ？（LCA 求两点距离）

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3252    Accepted Submission(s): 1222

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

 

Input

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

 

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

 

Sample Input

  

   2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1
  

 

Sample Output

  

   10
25
100
100
  

 

题意：给出一个无向带权图，给出m个询问a b，对于每个询问，求a到b的距离。

思路：参考了 http://www.cnblogs.com/ylfdrib/archive/2010/11/03/1867901.html。lca是离线算法，在求dfs的过程中把根节点到各节点的距离保存起来，最后对于每两点（a,b）之间的距离，d=dis[a]+dis[b]-2*dis[lca(a,b)]。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#include <cstdlib>
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
#define ll long long
using namespace std;

const int maxn=40005;

struct node
{
    int v,w,next;
}edge[maxn*2];
struct node1
{
    int u,v,next,id;
}query[maxn*2];
int G[maxn],Q[maxn],dis[maxn],fa[maxn],lca[500];
bool vis[maxn];
int n,m,s,e,num,numq,ans;
void init()
{
    memset(G,-1,sizeof(G));
    memset(Q,-1,sizeof(Q));
    num=numq=0;
}
void add(int u,int v,int w)
{
    edge[num].w=w;
    edge[num].v=v;
    edge[num].next=G[u];
    G[u]=num++;
}
void addq(int u,int v,int id)
{
    query[numq].u=u;
    query[numq].v=v;
    query[numq].id=id;
    query[numq].next=Q[u];
    Q[u]=numq++;
}
int Find_set(int x)
{
    return x==fa[x]?x:fa[x]=Find_set(fa[x]);
}
void input()
{
    int a,b,c;
    scanf("%d%d",&n,&m);
    for(int i=1;i<n;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);
        add(b,a,c);
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        addq(a,b,i);
        addq(b,a,i);
    }
}
void dfs(int u)
{
    vis[u]=true;
    fa[u]=u;
    for(int i=Q[u];i!=-1;i=query[i].next)
    {
        int v=query[i].v;
        if(vis[v]) lca[query[i].id]=Find_set(v);
    }
    for(int i=G[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!vis[v])
        {
            dis[v]=dis[u]+edge[i].w;
            dfs(v);
            fa[v]=u;
        }
    }
}
void solve()
{
    memset(vis,false,sizeof(vis));
    dis[1]=0;
    dfs(1);
    for(int i=0;i<numq;i+=2)
    {
        int u=query[i].u;
        int v=query[i].v;
        int id=query[i].id;
        printf("%d\n",dis[u]+dis[v]-2*dis[lca[id]]);
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        input();
        solve();
    }
    return 0;
}